Internal Constraints in Finite Elasticity: Manifolds or not

Abstract

An internal constraint for an elastic material is described either by a submanifold Open image in new window of the space of deformation gradients or by a submanifold Open image in new window of the space of symmetric strain tensors. It was proved in Podio-Guidugli and Vianello (Arch. Ration. Mech. Anal. 105(2):105–121, 2001) that the dimension of an isotropic constraint Open image in new window is always 8, or, equivalently, that the dimension of an isotropic constraint Open image in new window is always 5, rigidity and conformality being the only exceptions. Recently, this statement has been questioned in Carroll (Int. J. Eng. Sci. 47:1142–1148, 2009), where it is suggested that isotropic constraints might exist which do not conform to the above prescriptions. It is shown here that this is not the case, because the proposed counterxamples lack a proper manifold structure.

One should notice that the crucial steps are: (1) Open image in new window is a submanifold and, thus, has a tangent space at each point; (2) the identity tensor belongs to Open image in new window. Only if we neglect such requirements it is possible to construct examples of isotropic constraints (as algebraic varieties) which seemingly violate the results proved in [9].

Finally, how can we prove that Sph and Dev are the only proper subspaces of Sym which satisfy property (22) for all rotations?

The barehanded proof given in [9, Lemma 5.3] is perhaps a bit cumbersome and not very elegant. Indeed, the whole topic could be investigated within the framework of the theory of group actions on polynomials, for which one might refer to the (application-oriented) introduction given in [6, Chaps. XII and XIII]. Of course, this approach to the proof of Proposition 2 would not be self-contained with the additional disadvantage of requiring references to mathematical tools not so widely known in this context.

Here, however, I propose a new proof which, in my opinion, has the advantage of being reasonably simple and self-contained. Additionally, this approach is based on techniques which are commonly used in the literature about finite elasticity.

A preliminary technical lemma is useful.

Lemma 1

LetBandA∉Sph be a pair of symmetric tensors such that, for all rotationsQ,

Proof

We show that condition B∉Sph would lead to a contradiction, and this will suffice to prove the lemma.

Under the assumptions, and in view of the spectral decomposition theorem for symmetric tensors, we know that A has (at least) one characteristic space of dimension precisely 1. Let a be an eigenvector spanning such space. Since B is also assumed to be not spherical we can easily find at least one rotation \(\bar{ \mathbf {Q} }\) such that \(\bar{ \mathbf{a} }:=\bar { \mathbf {Q} }{ \mathbf{a} }\) is not among its eigenvectors. Let \(\bar{ \mathbf {A} }:=\bar{ \mathbf {Q} }{ \mathbf {A} }\bar{ \mathbf {Q} }^{\mathrm{T}}\), so that \(\bar{ \mathbf{a} }\) spans a one-dimensional characteristic space Open image in new window for \(\bar{ \mathbf {A} }\). We now notice that \(\bar{ \mathbf {A} }\) and Bcan not commute since, otherwise, B would map Open image in new window into itself, and, in particular, this would imply that \(\mathbf {B} \bar{ \mathbf{a} }=\mu\bar{ \mathbf{a} }\), for some μ, forcing \(\bar{ \mathbf{a} }\) to be an eigenvector of B. Thus,